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The order of a leaf node in a $B^{+}$ tree is the maximum number of children it can have. Suppose that block size is $1$ kilobytes, the child pointer takes $7$ bytes long and search field value takes $14$ bytes long. The order of the leaf node is _________.

1. $16$
2. $63$
3. $64$
4. $65$

data pointer = child pointer + search field value = 14+7 = 21B

so max number of children it can have = 1KB/21B = 1024/21 = 48.76 = 48(approx)

All other options are greater than the 48 can go with option 1 as asnwer

but 16 is much less than 48
are all options wrong?
i think 16 can be because i can fix my b+ tree to work with the order of 16 but where as for the other options i cannot have more than 48 because of the 48(size) constraint
oh,, okay
how it will be 16 means how 48 can connect to 16?
what is data (record pointer here) . B+tree leaf have only one child pointer
Official Key is 9 ie ALL CORRECT
by
$key size*(n-1)+n*child pointer \leqslant block size$

$14*(n-1)+7*n \leqslant 1024$

$21*n-14\leqslant1024$

$21*n\leqslant1038$

$n\leqslant1038/21$

$n\leqslant49.42$

$n$  should be less than $49.42$

i think ans should be Option A($16\leqslant49.42$)

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